There are two-digit numbers such that the sum of these digits is 6 while the product of the digits is 1/3 of the original number. Find the numbers.

Expert Answers
embizze eNotes educator| Certified Educator

Find all numbers of the form 10p+q, p and q nonnegative integers, such that p+q=6 and `pq=1/3(10p+q)` :

We can write all numbers of the form 10p+q such that p+q=6:


Now `1*5=1/3(15)`






The numbers we seek are 15 and 24


ayush01 | Student

here let the digit of ten's place be x and that of one's place be y. The the no. can be written in the form 10x+y. for eg: 91 can be given as 10*9+1=90+1=91.

now, x+y=6 and xy=1/3(10x+y)

                        or, xy=10x/3+y/3


                        or, x(y-10/3)=y/3

                        or, x=(y/3)/(y-10/3)

put value of x in x+y=6

then solve and you get,


Thus either y=5 OR y=4

when y=5, x=6-5=1

when y=6, x=6-4=2

Thus one no.=10*1+5=15

another no. =10*2+4=24

Therefore, the two numbers are 15 and 24.